Finding nth derivative of an exponential function and its value at the origin. I have a function defined as
$f(x) = e^{-\frac{1}{x^2}}, $if $ x\ne0$; $0$ if $x =0$.
where $f:[0,\infty) \to \mathbb{R}$ 
I am asked to prove the following: 
(a) that the nth derivative is of the form:
$f^{(n)}(x) =e^{-\frac{1}{x^2}}\frac{P_n (x)}{x^{N_n}}$
where $P_n (x)$ is a polynomial and $N_n \in \mathbb{N}$.
(b) $f^{(n)}(0) = 0$
(c) The Taylor Polynomials of $f$ around $0$ do not converge except on the point $x=0$.
For the point (a) I proceeded by induction over n, and used the fact that product and sum of polynomials is a polynomial, and so is the derivative of a polynomial.
However I am having some trouble with point (b) I tried to calculate
$\lim_{x\to 0}e^{-\frac{1}{x^2}}\frac{P_n (x)}{x^{N_n}}$
taking $x=\frac{1}{t}$ and calculating the limit as $t\to\infty$. But, because of the product rule I always have a $t^{N_n}P_n^{(n)} (x)$ and I don't know the degree of $P_n(x)$ (it gets pretty messy).
EDIT Am I a least on the right track? Or am I missing the point?
I would appreciate any hints and/or comments. Thank you for reading.
 A: Since
$(x^{-k})'
=(-k)x^{-k-1}
$,
he first terms
of $f^{(n)}(x)$
are
$f'(x)
=(-2x^{-3})e^{-x^{-2}}
=\frac{-2}{x^{3}}e^{-x^{-2}}
$
and
$f''(x)
=e^{-x^{-2}}(6x^{-4}+(-2x^{-3})^2)
=e^{-x^{-2}}(6x^{-4}+4x^{-6})
=e^{-x^{-2}}\frac{6x^2+4}{x^6}
$.
Suppose
$f^{(n)}(x)
=\frac{p_n(x)}{x^{cn}}e^{-x^{-2}}
=p_n(x)x^{-cn}e^{-x^{-2}}
$.
Since
$(p_n(x)x^{-cn})'
=p'_n(x)x^{-cn}-cnp_n(x)x^{-cn-1}
=(xp'_n(x)-cnp_n(x))x^{-cn-1}
$,
$\begin{array}\\
f^{(n+1)}(x)
&=(f^{(n)}(x))'\\
&=(p_n(x)x^{-cn}e^{-x^{-2}})'\\
&=(p_n(x)x^{-cn})'e^{-x^{-2}}+(p_n(x)x^{-cn})(e^{-x^{-2}})'\\
&=(p_n(x)x^{-cn})'e^{-x^{-2}}+(p_n(x)x^{-cn})(2x^{-3}e^{-x^{-2}})\\
&=e^{-x^{-2}}((p_n(x)x^{-cn})'+2p_n(x)x^{-cn-3})\\
&=e^{-x^{-2}}((xp'_n(x)-cnp_n(x))x^{-cn-1}+2p_n(x)x^{-cn-3})\\
&=e^{-x^{-2}}x^{-cn-3}((x^2(xp'_n(x)-cnp_n(x))+2p_n(x))\\
&=e^{-x^{-2}}x^{-cn-3}(x^3p'_n(x)-cnx^2p_n(x)+2p_n(x))\\
&=e^{-x^{-2}}x^{-cn-3}(x^3p'_n(x)-(cnx^2-2)p_n(x))\\
\end{array}
$
If there are no errors here
(prob < .7),
$c=3$ works
and then
$p_{n+1}(x)
=x^3p'_n(x)-(3nx^2-2)p_n(x)
$.
Note that
$p_{n+1}(0)
=2p_n(0)
$,
so all of the
$p_n(0)$
are non-zero
(and, in fact,
$p_n(0)=2^{n+1}$).
So
$f^{(n)}(x)
=p_n(x)x^{-3n}e^{-x^{-2}}
$.
Therefore
$\begin{array}\\
\lim_{x \to 0} f^{(n)}(x)
&=\lim_{x \to 0} p_n(x)x^{-3n}e^{-x^{-2}}\\
&=p_n(0)\lim_{x \to 0} p_n(0)x^{-3n}e^{-x^{-2}}\\
&=p_n(0)\lim_{x \to 0} e^{-x^{-2}-3n\ln(x)}\\
&=p_n(0)\lim_{x \to 0} e^{-x^{-2}(1+3nx^2\ln(x))}\\
\end{array}
$.
Since
$\lim_{x \to 0}x \ln x
=0
$
(setting
$x = 1/y$,
$x \ln x
=(1/y)\ln(1/y)
=-\ln(y^{1/y})
\to 0
$
since
$y^{1/y}
\to 1$
as
$y \to \infty$
.
Therefore
$\begin{array}\\
\lim_{x \to 0} f^{(n)}(x)
&=p_n(0)\lim_{x \to 0} e^{-x^{-2}(1+3nx^2\ln(x))}\\
&=p_n(0)\lim_{x \to 0} e^{-x^{-2}}\\
&= 0\\
\end{array}
$.
